MCAT Physics Question — New Force
- Nov 05, 2014
- MCAT Question of the Day
The force between the charges is given by Coulomb’s law:
F = kQ1q2 / r2
We may start by noting that changing one of the charges from positive to negative (or vice versa), as the question indicates when it says the valence on one of the charges is flipped, would immediately change the force from attractive to repulsive, or vice versa. Thus, we may immediately narrow our choices down to (B) and (C).
Next, if the distance between the charges in tripled and one charge in increased by a factor of 9, we have the following:
F = k(9Q1)q2 / (3r)2
F = 9kQ1q2 / (9)(r2)
F = kQ1q2 / r2
Thus we are left with a magnitue of force that is unchanged. Therefore, the answer is (B). Same magnitude, opposite direction.
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